Let the sum of the first “n” terms of a non-constant A.P.;\[{{a}_{1}},{{a}_{2}},{{a}_{3}},........\] be \[50n+\left( \dfrac{n(n-7)}{2} \right)\text{A}\] ,where A is a constant. If d is the common difference of this A.P. ; then the ordered pair (d,\[{{\text{a}}_{50}}\]) is equal to ?
(a). (A,50+46A)
(b). (A+50+45A)
(c). (50,50+46A)
(d). (50,50+45A)
Answer
606.9k+ views
Hint: It is easy to see that if we subtract the sum of first “50 terms”, to “first 49 terms” then we will get the 50-th term of this Arithmetic Progression. This is the key idea of the problem. In order to determine the common difference, we will use the same idea.
Complete step-by-step answer:
Let us denote the sum of first “n” terms is \[{{S}_{n}}\]. In particular, we will find the value when n=50,49,2 and 1 to get the ordered pair.
So,
\[{{S}_{n}}=50n+\left( \dfrac{n(n-7)}{2} \right)\text{A}............................\text{(1)}\]
We will first calculate, n=50
\[{{S}_{50}}=50.50+\left( \dfrac{50(50-7)}{2} \right)A\]
\[\Rightarrow {{S}_{50}}=2500+1075A...............................(2)\]
Then, we will calculate, n=49;
\[{{S}_{49}}=50.49+\left( \dfrac{49(49-7)}{2} \right)A\]
\[\Rightarrow {{S}_{49}}=2450+1029A.................................(3)\]
Then we will subtract (3) from (2) to get the 50th term.
\[{{a}_{50}}=50+46A\].
This is the 2nd half of the ordered pair.
For 1st part we need to calculate \[{{S}_{2}}\And {{S}_{1}}\]
\[{{S}_{2}}=50.2+\left( \dfrac{2(2-7)}{2} \right)A\]
\[\Rightarrow {{S}_{2}}=100-5A\]……………………..(3)
\[{{S}_{1}}=50.1+\left( \dfrac{1(1-7)}{2} \right)A\]
\[\Rightarrow {{S}_{1}}=50-3A\]……………………(4)
Actually (4) denotes the first term of the arithmetic progression. So, the 2nd term of the progression will be the result of subtraction from (3) to (2).
\[{{a}_{2}}=100-5A-50+3A\]
\[\Rightarrow {{a}_{2}}=50-2A\]
So, the common difference will be,
\[d=50-2A-50+3A\]
\[\Rightarrow d=A\]
So, our final answer will be (A,50+46A).
So, the correct answer is “Option A”.
Note: Please do not try to use the formula for the arithmetic progression. As in question, there is no given information about the first term and the common difference. Be careful about the word “ordered pair” mentioned in the question as it indicates that order is important. There is an alternative method to find the 50th element of the arithmetic progression.
Complete step-by-step answer:
Let us denote the sum of first “n” terms is \[{{S}_{n}}\]. In particular, we will find the value when n=50,49,2 and 1 to get the ordered pair.
So,
\[{{S}_{n}}=50n+\left( \dfrac{n(n-7)}{2} \right)\text{A}............................\text{(1)}\]
We will first calculate, n=50
\[{{S}_{50}}=50.50+\left( \dfrac{50(50-7)}{2} \right)A\]
\[\Rightarrow {{S}_{50}}=2500+1075A...............................(2)\]
Then, we will calculate, n=49;
\[{{S}_{49}}=50.49+\left( \dfrac{49(49-7)}{2} \right)A\]
\[\Rightarrow {{S}_{49}}=2450+1029A.................................(3)\]
Then we will subtract (3) from (2) to get the 50th term.
\[{{a}_{50}}=50+46A\].
This is the 2nd half of the ordered pair.
For 1st part we need to calculate \[{{S}_{2}}\And {{S}_{1}}\]
\[{{S}_{2}}=50.2+\left( \dfrac{2(2-7)}{2} \right)A\]
\[\Rightarrow {{S}_{2}}=100-5A\]……………………..(3)
\[{{S}_{1}}=50.1+\left( \dfrac{1(1-7)}{2} \right)A\]
\[\Rightarrow {{S}_{1}}=50-3A\]……………………(4)
Actually (4) denotes the first term of the arithmetic progression. So, the 2nd term of the progression will be the result of subtraction from (3) to (2).
\[{{a}_{2}}=100-5A-50+3A\]
\[\Rightarrow {{a}_{2}}=50-2A\]
So, the common difference will be,
\[d=50-2A-50+3A\]
\[\Rightarrow d=A\]
So, our final answer will be (A,50+46A).
So, the correct answer is “Option A”.
Note: Please do not try to use the formula for the arithmetic progression. As in question, there is no given information about the first term and the common difference. Be careful about the word “ordered pair” mentioned in the question as it indicates that order is important. There is an alternative method to find the 50th element of the arithmetic progression.
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